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7h
comment Closed form for $\sum^\infty_{n=1}\frac{H_n}{2^n\,(2n+1)^2}$
Related questions: (1), (2).
8h
comment Closed form for $\sum^\infty_{n=1}\frac{H_n}{2^n\,(2n+1)^2}$
Using an integral representation $H_n=\int_0^1\frac{1-x^n}{1-x}dx$ the series $S$ can be converted to an integral containing dilogarithms, that can be evaluated and simplified using Mathematica to this form that numerically matches my conjectured result $(2)$ and contains similar terms.
9h
asked Closed form for $\sum^\infty_{n=1}\frac{H_n}{2^n\,(2n+1)^2}$
Aug
28
awarded  Good Question
Aug
27
accepted A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
Aug
27
comment A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
I found some useful results for harmonic numbers here: pi314.net/eng/hypergse13.php
Aug
27
awarded  Nice Question
Aug
26
comment A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
$\mathcal S_{14}\stackrel?=3\left(\ln^2\sin\frac\pi7+\ln^2\sin\frac{2\pi}7+\ln^2\sin\frac{‌​3\pi}7\right) - 8\ln^22 - \frac14\ln^27 + 4\ln2\cdot\ln7 - \frac{25}{84}\pi^2$. It seems that some pattern starts to emerge...
Aug
26
comment A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
$\mathcal S_{16}\stackrel?=\frac12 \ln^2 2+\ln^2 \cos\frac\pi{16} + \ln^2 \cos\frac{3\pi}{16} + \ln^2 \sin\frac\pi{16} + \ln^2\sin\frac{3\pi}{16} - \frac{65}{192}\pi^2$.
Aug
26
comment A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
Great answer! Have you been able to simplify the sum of dilogarithms of complex arguments for $m=5$?
Aug
26
comment A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
And $\mathcal S_{10}\stackrel?=3\ln^2 2+2\ln^2(1+\sqrt5)+\ln2\cdot\ln5-4\ln2\cdot\ln(1+\sqrt5)-\frac{13}{60}\pi^2$.
Aug
26
comment A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
Also, $\mathcal S_8\stackrel?=\frac{21}8\ln^2 2+\frac12\ln^2(1+\sqrt2)-\frac{17}{96}\pi^2$ and $\mathcal S_{12}\stackrel?=\frac74\ln^2 2+\ln2\cdot\ln3+\frac12\ln^2(2+\sqrt3)-\frac{37}{144}\pi^2$.
Aug
26
comment A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
I remembered that there was a closed form for $\operatorname{Li}_2(\phi^{-2})$, but for some reason I forgot that $\operatorname{Li}_2(\phi^{-1})$ had it as well, and did not even check. Thanks for the reminder!
Aug
26
revised A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
added 174 characters in body
Aug
26
asked A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
Aug
23
awarded  Nice Question
Aug
23
accepted A proof that $\frac{(2\phi)^n-(-1)^n}{\phi^{2n}-(-1)^n}\cdot\left(2^n-\phi^n\right)\cdot\sqrt5\in\mathbb Q$ for all $n\in\mathbb Z$
Aug
23
asked A proof that $\frac{(2\phi)^n-(-1)^n}{\phi^{2n}-(-1)^n}\cdot\left(2^n-\phi^n\right)\cdot\sqrt5\in\mathbb Q$ for all $n\in\mathbb Z$
Aug
22
comment Conjectured value of a harmonic sum $\sum_{n=1}^\infty\left(H_n-\,2H_{2n}+H_{4n}\right)^2$
One more simple series where $O(n^{-1})$ term is cancelled using harmonic numbers only: $\sum_{n=1}^\infty\left(H_n-4H_{2n}+5H_{4n}-2H_{8n}\right)=\frac\pi{4\sqrt2} - \frac{3\pi}{16}.$
Aug
22
comment Conjectured value of a harmonic sum $\sum_{n=1}^\infty\left(H_n-\,2H_{2n}+H_{4n}\right)^2$
And another: $\sum_{n=1}^\infty\frac{(-1)^n}n\left(H_n-\,2H_{2n}+H_{4n}\right)=\frac12 \ln^2\!\left(1+\sqrt2\right)+\frac18 \ln^2 2-\frac{5\pi^2}{96}.$